# The continued fraction expansion for e

• Oct 16th 2006, 11:01 AM
beta12
The continued fraction expansion for e
The continued fraction expansion for the number e = 2.718281828459045.....
is [2;1,2,1,1,4,1,1,6,1,1,8,1,1,...].

How many terms are required to get 4 decimal places (2.718) correct?

Please teach me how to solve this question. Thank you very much.
• Oct 16th 2006, 06:14 PM
ThePerfectHacker
Quote:

Originally Posted by beta12
The continued fraction expansion for the number e = 2.718281828459045.....
is [2;1,2,1,1,4,1,1,6,1,1,8,1,1,...].

How many terms are required to get 4 decimal places (2.718) correct?

That is a classical expansion by Euler (sadly not even I know the proof).

But I hope you are asking not to prove it but to get 4 decimals?

Using a well-known inequlality from continued fractions:
|x-p_n/q_n|<=(1/q_n)^2
In order to get 4 decimal points we require that,
(1/q_n)^2<=.0001
Equivalently,
1/q_n<=.0316
Thus,
q_n>=31.62
Thus,
q_n>=32
That is the smallest "n" that makes the denominator of the convergent at least 32.
• Oct 17th 2006, 06:22 AM
beta12
Hi Perfecthacker,

Thank you very much.